Question

Find the equation of the circle which passes through the points (-3,0), (3,2) and (5,4)​

Answer 1

Answer:

$(x + 3)^{2} + (y - 10)^{2} = 10^{2}$.

Step-by-step explanation:

There are three quantities in the equation of a circle in a plane:

• $x$-coordinate of the center of the circle,
• $y$-coordinate of the center of the circle, and
• radius of the circle.

Let $(a,\, b)$ denote the center of this circle and let $r$ denote the radius of this circle ($r > 0$.) The equation of this circle would be:

$(x - a)^{2} + (y - b)^{2} = r^{2}$.

A point $(x_{1},\, y_{1})$ is on this circle if and only if the equation of this circle is satisfied after substituting in $x = x_{1}$ and $y = y_{1}$.

For example, since $(-3,\, 0)$ is on this circle, the equation of this circle should continue to hold after substituting in $x = (-3)$ and $y = 0$:

$((-3) - a)^{2} + (0 - b)^{2} = r^{2}$.

Likewise, for the other two given points that are on this circle, the following would be equations:

$(3 - a)^{2} + (2 - b)^{2} = r^{2}$, and

$(5 - a)^{2} + (4 - b)^{2} = r^{2}$.

Expand each of these equations using the binomial theorem.

$9 + 6\, a + a^{2} + b^{2} = r^{2}$ (note that $((-3) - a)^{2}$ is equivalent to $(3 + a)^{2}$.)

$9 - 6\, a + a^{2} + 4 - 4\, b + b^{2} = r^{2}$.

$25 - 10\, a + a^{2} + 16 - 8\, b + b^{2} = r^{2}$.

These three equations are part of a system of three equations and three unknowns ($a$, $b$, and $r$ with $r > 0$.)

Subtract the second equation from the first to eliminate the non-linear terms ($a^{2}$, $b^{2}$, and $r^{2}$) and simplify to obtain a linear equation about $a$ and $b$:

$3\, a + b = 1$.

Likewise, subtract the third equation from the first and simplify to obtain:

$2\, a + b = 4$.

Solve the system of these two equations for $a$ and $b$:

$a = -3$.

$b = 10$.

Substitute $a = -3$ and $b = 10$ into any one of the three original equations that include $r$ and solve for $r\!$. For example, substituting into the first equation to obtain:

$\begin{aligned}r^{2} &= 9 + 6 \times (-3) + (-3)^{2} + 10^{2} \\ &= 10^{2}\end{aligned}$.

Since $r > 0$, $r = 10$ would be the only solution to this system of equations.

Therefore, the equation of this circle would be $(x - (-3))^{2} + (y - 10)^{2} = 10^{2}$.

Simplify to obtain: $(x + 3)^{2} + (y - 10)^{2} = 10^{2}$.